= Kepler's second law
{c}
{title2=$dA/dt=h/2$}
The relative displacement of a <Kepler orbit> sweeps equal areas about the attracting <focus of a conic> in equal times. A short interval near <periapsis> covers a wide angle at small separation; the same interval near <apoapsis> covers a narrow angle at large separation. The swept areas are equal even though the travelled distances are different. In particular, the sectors have curved <ellipse> boundaries: triangles formed by joining their endpoints only approximate the swept areas for short intervals.
Let $h=|\mathbf r\times\dot{\mathbf r}|$ be the relative <specific angular momentum>. The <areal velocity> is
$$
\frac{dA}{dt}=\frac12|\mathbf r\times\dot{\mathbf r}|=\frac h2.
$$
A <central force> exerts zero <torque> about its centre, so <conservation of angular momentum> makes this rate constant. This explanation also applies to other <central forces>; the <ellipse> shape specifically comes from <Newtonian gravity>.
In the $a=1$, $e=0.6$, $\mu=1$ example, <specific angular momentum> is $h=\sqrt{\mu a(1-e^2)}=\sqrt{0.64}=0.8$, giving <areal velocity> $0.4$. At <periapsis> and <apoapsis>, the <velocity> is perpendicular to the relative displacement, so $h=rv$. The endpoint <speeds> are consequently
$$
v_{\rm p}=\frac{0.8}{0.4}=2,\qquad v_{\rm a}=\frac{0.8}{1.6}=0.5.
$$
The fourfold <speed> difference is consistent with the fourfold separation difference. Away from these endpoints, $h/r$ gives only the transverse component of <velocity>; the full <speed> follows from the <vis-viva equation>.
To actually place equal-time markers, use <Kepler's equation>. Write <mean anomaly> as $\mathcal M=nt$, with $t=0$ at <periapsis> and <mean motion> $n=\sqrt{\mu/a^3}=1$. The different symbol $\mathcal M$ avoids confusing this angle with the <mass> $M$. Solve for <eccentric anomaly> $E$ in <radians>, then compute the relative position:
$$
\mathcal M=E-e\sin E,\qquad x=a(\cos E-e),\qquad y=b\sin E.
$$
This timing procedure is explained in https://pwg.gsfc.nasa.gov/stargaze/Smotion.htm[NASA educational archive, How Orbital Motion is Calculated].
\Image[/kepler-equal-time-areas.png]
{title=Equal-time positions and equal swept areas on the relative ellipse with a=1 and e=0.6}
For the first interval $0\le t\le\pi/6$, the final <eccentric anomaly> is $E\approx1.041495$. For the equal interval $\pi\le t\le7\pi/6$ starting at <apoapsis>, it ends at $E\approx3.471065$. Both swept areas are $\Delta A=(h/2)\Delta t=0.4\pi/6=\pi/15\approx0.209440$, one twelfth of the entire <ellipse>'s area $0.8\pi$. The diagram computes its markers from <Kepler's equation>, making visible how uniform time steps become uneven spatial steps.
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